Q. WHAT IS CUSTOMER CHURNING?

When consumers or subscribers stop doing business with a company or service, this is known as customer churn. Customers in the telecom industry have a number of service providers from which to choose and can actively switch from one to the next. In this extremely competitive sector, the telecoms industry has an annual turnover rate of 15-25 percent. Because most businesses have a huge number of clients and can't afford to spend much time with each of them, personalized customer retention is difficult. The additional money would be insufficient to cover the costs. If a company could predict which customers are likely to depart ahead of time, it might focus customer retention efforts only on these "high risk" consumers.

Q. How is model used to predict customer churning?

The major goal of the customer churn prediction model is to proactively engage with customers who are most likely to churn. Offer a gift certificate or special price and lock them in for another year or two to increase their lifetime worth to the firm.

There are two broad concepts to understand here:

  1. We want a client churn prediction models that can forecast attrition in advance (say, one month, three months, or even six months ahead of time – it all depends on the use-case). This implies you must be very careful with the cut-off date, i.e. you should not include any information after the cut-off date as a feature in the machine learning model, since this would result in leakage. The Event is the time period preceding the cut-off date.
  2. In most cases, you'll have to work a little bit to establish a target column for customer churn prediction; it's usually not accessible in the form you'd want. If you want to anticipate if a client would churn in the next quarter, for example, you'll go through all of your current customers as of your event cut-off date and see if they left the firm within next quarter or not (1 for yes, 0 for no). Productivity Period is the name of the quarter in this scenario.

HOW TO REDUCE CUSTOMER CHURINING?

To reduce customer churn, telecom companies need to predict which customers are at high risk of churn. To detect early signs of potential churn, one must first develop a holistic view of the customers and their interactions across numerous channels, including store/branch visits, product purchase histories, customer service calls, Web-based transactions, and social media interactions, to mention a few. To detect early indicators of possible churn, create a holistic picture of consumers and their interactions across many channels, such as store/branch visits, product purchase histories, customer care calls, Web-based transactions, and social media interactions, to name a few. As a consequence, managing churn may help these companies not only maintain their market position, but also expand and prosper. The lower the cost of initiation and the higher the profit, the more clients they have in their network. As a result, the company's primary focus for success is on lowering client attrition and developing a successful retention plan.

STARTING WITH IMPORTING ALL THE IMPORTANT LIBRARIES.

In [3]:
from sklearn.preprocessing import StandardScaler
from sklearn.preprocessing import LabelEncoder
In [7]:
from sklearn.tree import DecisionTreeClassifier
from sklearn.ensemble import RandomForestClassifier
from sklearn.linear_model import LogisticRegression
from sklearn.metrics import roc_curve
In [8]:
from sklearn.model_selection import train_test_split
from sklearn.metrics import accuracy_score
from sklearn.metrics import recall_score, confusion_matrix, precision_score, f1_score, accuracy_score, classification_report
In [9]:
df = pd.read_csv('D:/Python/ML/WA_Fn-UseC_-Telco-Customer-Churn.csv')
In [11]:
df.head()
Out[11]:
customerID gender SeniorCitizen Partner Dependents tenure PhoneService MultipleLines InternetService OnlineSecurity ... DeviceProtection TechSupport StreamingTV StreamingMovies Contract PaperlessBilling PaymentMethod MonthlyCharges TotalCharges Churn
0 7590-VHVEG Female 0 Yes No 1 No No phone service DSL No ... No No No No Month-to-month Yes Electronic check 29.85 29.85 No
1 5575-GNVDE Male 0 No No 34 Yes No DSL Yes ... Yes No No No One year No Mailed check 56.95 1889.5 No
2 3668-QPYBK Male 0 No No 2 Yes No DSL Yes ... No No No No Month-to-month Yes Mailed check 53.85 108.15 Yes
3 7795-CFOCW Male 0 No No 45 No No phone service DSL Yes ... Yes Yes No No One year No Bank transfer (automatic) 42.30 1840.75 No
4 9237-HQITU Female 0 No No 2 Yes No Fiber optic No ... No No No No Month-to-month Yes Electronic check 70.70 151.65 Yes

5 rows × 21 columns

In [12]:
df.shape
Out[12]:
(7043, 21)
In [13]:
df.info()

<class 'pandas.core.frame.DataFrame'>
RangeIndex: 7043 entries, 0 to 7042
Data columns (total 21 columns):
 #   Column            Non-Null Count  Dtype  
---  ------            --------------  -----  
 0   customerID        7043 non-null   object 
 1   gender            7043 non-null   object 
 2   SeniorCitizen     7043 non-null   int64  
 3   Partner           7043 non-null   object 
 4   Dependents        7043 non-null   object 
 5   tenure            7043 non-null   int64  
 6   PhoneService      7043 non-null   object 
 7   MultipleLines     7043 non-null   object 
 8   InternetService   7043 non-null   object 
 9   OnlineSecurity    7043 non-null   object 
 10  OnlineBackup      7043 non-null   object 
 11  DeviceProtection  7043 non-null   object 
 12  TechSupport       7043 non-null   object 
 13  StreamingTV       7043 non-null   object 
 14  StreamingMovies   7043 non-null   object 
 15  Contract          7043 non-null   object 
 16  PaperlessBilling  7043 non-null   object 
 17  PaymentMethod     7043 non-null   object 
 18  MonthlyCharges    7043 non-null   float64
 19  TotalCharges      7043 non-null   object 
 20  Churn             7043 non-null   object 
dtypes: float64(1), int64(2), object(18)
memory usage: 1.1+ MB
In [14]:
df.columns.values
Out[14]:
array(['customerID', 'gender', 'SeniorCitizen', 'Partner', 'Dependents',
       'tenure', 'PhoneService', 'MultipleLines', 'InternetService',
       'OnlineSecurity', 'OnlineBackup', 'DeviceProtection',
       'TechSupport', 'StreamingTV', 'StreamingMovies', 'Contract',
       'PaperlessBilling', 'PaymentMethod', 'MonthlyCharges',
       'TotalCharges', 'Churn'], dtype=object)
In [15]:
df.dtypes
Out[15]:
customerID           object
gender               object
SeniorCitizen         int64
Partner              object
Dependents           object
tenure                int64
PhoneService         object
MultipleLines        object
InternetService      object
OnlineSecurity       object
OnlineBackup         object
DeviceProtection     object
TechSupport          object
StreamingTV          object
StreamingMovies      object
Contract             object
PaperlessBilling     object
PaymentMethod        object
MonthlyCharges      float64
TotalCharges         object
Churn                object
dtype: object

THE TARGET OR DEPENDENT VARIABLE IS "CHURN"

Visualize missing values as a matrix

In [17]:
msno.matrix(df);

We can rapidly detect the missingness trend in the dataset to use this matrix.

We can see from the above visualization that there is no unusual pattern that sticks out. In truth, there is no data that is missing.

DATA MANIPULATION

In [18]:
df = df.drop(['customerID'], axis = 1)
df.head()
Out[18]:
gender SeniorCitizen Partner Dependents tenure PhoneService MultipleLines InternetService OnlineSecurity OnlineBackup DeviceProtection TechSupport StreamingTV StreamingMovies Contract PaperlessBilling PaymentMethod MonthlyCharges TotalCharges Churn
0 Female 0 Yes No 1 No No phone service DSL No Yes No No No No Month-to-month Yes Electronic check 29.85 29.85 No
1 Male 0 No No 34 Yes No DSL Yes No Yes No No No One year No Mailed check 56.95 1889.5 No
2 Male 0 No No 2 Yes No DSL Yes Yes No No No No Month-to-month Yes Mailed check 53.85 108.15 Yes
3 Male 0 No No 45 No No phone service DSL Yes No Yes Yes No No One year No Bank transfer (automatic) 42.30 1840.75 No
4 Female 0 No No 2 Yes No Fiber optic No No No No No No Month-to-month Yes Electronic check 70.70 151.65 Yes
In [19]:
df['TotalCharges'] = pd.to_numeric(df.TotalCharges, errors='coerce')
df.isnull().sum()
Out[19]:
gender               0
SeniorCitizen        0
Partner              0
Dependents           0
tenure               0
PhoneService         0
MultipleLines        0
InternetService      0
OnlineSecurity       0
OnlineBackup         0
DeviceProtection     0
TechSupport          0
StreamingTV          0
StreamingMovies      0
Contract             0
PaperlessBilling     0
PaymentMethod        0
MonthlyCharges       0
TotalCharges        11
Churn                0
dtype: int64

Here we see that the TotalCharges has 11 missing values. Let's check this data.

In [20]:
df[np.isnan(df['TotalCharges'])]
Out[20]:
gender SeniorCitizen Partner Dependents tenure PhoneService MultipleLines InternetService OnlineSecurity OnlineBackup DeviceProtection TechSupport StreamingTV StreamingMovies Contract PaperlessBilling PaymentMethod MonthlyCharges TotalCharges Churn
488 Female 0 Yes Yes 0 No No phone service DSL Yes No Yes Yes Yes No Two year Yes Bank transfer (automatic) 52.55 NaN No
753 Male 0 No Yes 0 Yes No No No internet service No internet service No internet service No internet service No internet service No internet service Two year No Mailed check 20.25 NaN No
936 Female 0 Yes Yes 0 Yes No DSL Yes Yes Yes No Yes Yes Two year No Mailed check 80.85 NaN No
1082 Male 0 Yes Yes 0 Yes Yes No No internet service No internet service No internet service No internet service No internet service No internet service Two year No Mailed check 25.75 NaN No
1340 Female 0 Yes Yes 0 No No phone service DSL Yes Yes Yes Yes Yes No Two year No Credit card (automatic) 56.05 NaN No
3331 Male 0 Yes Yes 0 Yes No No No internet service No internet service No internet service No internet service No internet service No internet service Two year No Mailed check 19.85 NaN No
3826 Male 0 Yes Yes 0 Yes Yes No No internet service No internet service No internet service No internet service No internet service No internet service Two year No Mailed check 25.35 NaN No
4380 Female 0 Yes Yes 0 Yes No No No internet service No internet service No internet service No internet service No internet service No internet service Two year No Mailed check 20.00 NaN No
5218 Male 0 Yes Yes 0 Yes No No No internet service No internet service No internet service No internet service No internet service No internet service One year Yes Mailed check 19.70 NaN No
6670 Female 0 Yes Yes 0 Yes Yes DSL No Yes Yes Yes Yes No Two year No Mailed check 73.35 NaN No
6754 Male 0 No Yes 0 Yes Yes DSL Yes Yes No Yes No No Two year Yes Bank transfer (automatic) 61.90 NaN No

It can also be noted that the Tenure column is 0 for these entries even though the MonthlyCharges column is not empty. Let's see if there are any other 0 values in the tenure column.

In [21]:
df[df['tenure'] == 0].index
Out[21]:
Int64Index([488, 753, 936, 1082, 1340, 3331, 3826, 4380, 5218, 6670, 6754], dtype='int64')

There are no additional missing values in the Tenure column. Let's delete the rows with missing values in Tenure columns since there are only 11 rows and deleting them will not affect the data.

In [22]:
df.drop(labels=df[df['tenure'] == 0].index, axis=0, inplace=True)
df[df['tenure'] == 0].index
Out[22]:
Int64Index([], dtype='int64')

To solve the problem of missing values in TotalCharges column, I decided to fill it with the mean of TotalCharges values.

In [23]:
df.fillna(df["TotalCharges"].mean())
Out[23]:
gender SeniorCitizen Partner Dependents tenure PhoneService MultipleLines InternetService OnlineSecurity OnlineBackup DeviceProtection TechSupport StreamingTV StreamingMovies Contract PaperlessBilling PaymentMethod MonthlyCharges TotalCharges Churn
0 Female 0 Yes No 1 No No phone service DSL No Yes No No No No Month-to-month Yes Electronic check 29.85 29.85 No
1 Male 0 No No 34 Yes No DSL Yes No Yes No No No One year No Mailed check 56.95 1889.50 No
2 Male 0 No No 2 Yes No DSL Yes Yes No No No No Month-to-month Yes Mailed check 53.85 108.15 Yes
3 Male 0 No No 45 No No phone service DSL Yes No Yes Yes No No One year No Bank transfer (automatic) 42.30 1840.75 No
4 Female 0 No No 2 Yes No Fiber optic No No No No No No Month-to-month Yes Electronic check 70.70 151.65 Yes
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
7038 Male 0 Yes Yes 24 Yes Yes DSL Yes No Yes Yes Yes Yes One year Yes Mailed check 84.80 1990.50 No
7039 Female 0 Yes Yes 72 Yes Yes Fiber optic No Yes Yes No Yes Yes One year Yes Credit card (automatic) 103.20 7362.90 No
7040 Female 0 Yes Yes 11 No No phone service DSL Yes No No No No No Month-to-month Yes Electronic check 29.60 346.45 No
7041 Male 1 Yes No 4 Yes Yes Fiber optic No No No No No No Month-to-month Yes Mailed check 74.40 306.60 Yes
7042 Male 0 No No 66 Yes No Fiber optic Yes No Yes Yes Yes Yes Two year Yes Bank transfer (automatic) 105.65 6844.50 No

7032 rows × 20 columns

In [24]:
df.isnull().sum()
Out[24]:
gender              0
SeniorCitizen       0
Partner             0
Dependents          0
tenure              0
PhoneService        0
MultipleLines       0
InternetService     0
OnlineSecurity      0
OnlineBackup        0
DeviceProtection    0
TechSupport         0
StreamingTV         0
StreamingMovies     0
Contract            0
PaperlessBilling    0
PaymentMethod       0
MonthlyCharges      0
TotalCharges        0
Churn               0
dtype: int64
In [25]:
df["SeniorCitizen"]= df["SeniorCitizen"].map({0: "No", 1: "Yes"})
df.head()
Out[25]:
gender SeniorCitizen Partner Dependents tenure PhoneService MultipleLines InternetService OnlineSecurity OnlineBackup DeviceProtection TechSupport StreamingTV StreamingMovies Contract PaperlessBilling PaymentMethod MonthlyCharges TotalCharges Churn
0 Female No Yes No 1 No No phone service DSL No Yes No No No No Month-to-month Yes Electronic check 29.85 29.85 No
1 Male No No No 34 Yes No DSL Yes No Yes No No No One year No Mailed check 56.95 1889.50 No
2 Male No No No 2 Yes No DSL Yes Yes No No No No Month-to-month Yes Mailed check 53.85 108.15 Yes
3 Male No No No 45 No No phone service DSL Yes No Yes Yes No No One year No Bank transfer (automatic) 42.30 1840.75 No
4 Female No No No 2 Yes No Fiber optic No No No No No No Month-to-month Yes Electronic check 70.70 151.65 Yes
In [26]:
df["InternetService"].describe(include=['object', 'bool'])
Out[26]:
count            7032
unique              3
top       Fiber optic
freq             3096
Name: InternetService, dtype: object
In [27]:
numerical_cols = ['tenure', 'MonthlyCharges', 'TotalCharges']
df[numerical_cols].describe()
Out[27]:
tenure MonthlyCharges TotalCharges
count 7032.000000 7032.000000 7032.000000
mean 32.421786 64.798208 2283.300441
std 24.545260 30.085974 2266.771362
min 1.000000 18.250000 18.800000
25% 9.000000 35.587500 401.450000
50% 29.000000 70.350000 1397.475000
75% 55.000000 89.862500 3794.737500
max 72.000000 118.750000 8684.800000

DATA VISUALIZATION

In [28]:
s_labels = ['Male', 'Female']
I_labels = ['No', 'Yes']
# Create subplots: use 'domain' type for Pie subplot
fig = make_subplots(rows=1, cols=2, specs=[[{'type':'domain'}, {'type':'domain'}]])
fig.add_trace(go.Pie(labels=s_labels, values=df['gender'].value_counts(), name="Gender"),
              1, 1)
fig.add_trace(go.Pie(labels=I_labels, values=df['Churn'].value_counts(), name="Churn"),
              1, 2)

# Use `hole` to create a donut-like pie chart
fig.update_traces(hole=.4, hoverinfo="label+percent+name", textfont_size=16)

fig.update_layout(
    title_text="Gender and Churn Distributions",
    # Add annotations in the center of the donut pies.
    annotations=[dict(text='Gender', x=0.16, y=0.5, font_size=20, showarrow=False),
                 dict(text='Churn', x=0.84, y=0.5, font_size=20, showarrow=False)])
fig.show()

INTERPRETATION

26.6 % of customers switched to another firm. Customers are 49.5 % female and 50.5 % male.

In [29]:
df["Churn"][df["Churn"]=="No"].groupby(by=df["gender"]).count()
Out[29]:
gender
Female    2544
Male      2619
Name: Churn, dtype: int64
In [30]:
df["Churn"][df["Churn"]=="Yes"].groupby(by=df["gender"]).count()
Out[30]:
gender
Female    939
Male      930
Name: Churn, dtype: int64
In [33]:
plt.figure(figsize=(6, 6))
labels =["Churn: Yes","Churn:No"]
values = [1869,5163]
labels_gender = ["F","M","F","M"]
sizes_gender = [939,930 , 2544,2619]
colors = ['#ff6666', '#66b3ff']
colors_gender = ['#c2c2f0','#ffb3e6', '#c2c2f0','#ffb3e6']
explode = (0.3,0.3) 
explode_gender = (0.1,0.1,0.1,0.1)
textprops = {"fontsize":15}
#Plot
plt.pie(values, labels=labels,autopct='%1.1f%%',pctdistance=1.08, labeldistance=0.8,colors=colors, startangle=90,frame=True, explode=explode,radius=10, textprops =textprops, counterclock = True, )
plt.pie(sizes_gender,labels=labels_gender,colors=colors_gender,startangle=90, explode=explode_gender,radius=7, textprops =textprops, counterclock = True, )
#Draw circle
centre_circle = plt.Circle((0,0),5,color='black', fc='white',linewidth=0)
fig = plt.gcf()
fig.gca().add_artist(centre_circle)

plt.title('Churn Distribution w.r.t Gender: Male(M), Female(F)', fontsize=15, y=1.1)

# show plot 
 
plt.axis('equal')
plt.tight_layout()
plt.show()

There is negligible difference in customer percentage/ count who chnaged the service provider. Both genders behaved in similar fashion when it comes to migrating to another service provider/firm.

In [34]:
fig = px.histogram(df, x="Churn", color="Contract", barmode="group", title="<b>Customer contract distribution<b>")
fig.update_layout(width=700, height=500, bargap=0.1)
fig.show()

About 75% of customer with Month-to-Month Contract opted to move out as compared to 13% of customrs with One Year Contract and 3% with Two Year Contract

In [35]:
labels = df['PaymentMethod'].unique()
values = df['PaymentMethod'].value_counts()

fig = go.Figure(data=[go.Pie(labels=labels, values=values, hole=.3)])
fig.update_layout(title_text="<b>Payment Method Distribution</b>")
fig.show()
In [36]:
sns.set_context("paper",font_scale=1.1)
ax = sns.kdeplot(df.MonthlyCharges[(df["Churn"] == 'No') ],
                color="Red", shade = True);
ax = sns.kdeplot(df.MonthlyCharges[(df["Churn"] == 'Yes') ],
                ax =ax, color="Blue", shade= True);
ax.legend(["Not Churn","Churn"],loc='upper right');
ax.set_ylabel('Density');
ax.set_xlabel('Monthly Charges');
ax.set_title('Distribution of monthly charges by churn');

Customers with higher Monthly Charges are also more likely to churn

In [37]:
fig = px.box(df, x='Churn', y = 'tenure')

# Update yaxis properties
fig.update_yaxes(title_text='Tenure (Months)', row=1, col=1)
# Update xaxis properties
fig.update_xaxes(title_text='Churn', row=1, col=1)

# Update size and title
fig.update_layout(autosize=True, width=750, height=600,
    title_font=dict(size=25, family='Courier'),
    title='<b>Tenure vs Churn</b>',
)

fig.show()

New customers are more likely to churn

DATA PREPROCESSING

In [38]:
def object_to_int(dataframe_series):
    if dataframe_series.dtype=='object':
        dataframe_series = LabelEncoder().fit_transform(dataframe_series)
    return dataframe_series
In [39]:
df = df.apply(lambda x: object_to_int(x))
df.head()
Out[39]:
gender SeniorCitizen Partner Dependents tenure PhoneService MultipleLines InternetService OnlineSecurity OnlineBackup DeviceProtection TechSupport StreamingTV StreamingMovies Contract PaperlessBilling PaymentMethod MonthlyCharges TotalCharges Churn
0 0 0 1 0 1 0 1 0 0 2 0 0 0 0 0 1 2 29.85 29.85 0
1 1 0 0 0 34 1 0 0 2 0 2 0 0 0 1 0 3 56.95 1889.50 0
2 1 0 0 0 2 1 0 0 2 2 0 0 0 0 0 1 3 53.85 108.15 1
3 1 0 0 0 45 0 1 0 2 0 2 2 0 0 1 0 0 42.30 1840.75 0
4 0 0 0 0 2 1 0 1 0 0 0 0 0 0 0 1 2 70.70 151.65 1
In [40]:
plt.figure(figsize=(14,7))
df.corr()['Churn'].sort_values(ascending = False)
Out[40]:
Churn               1.000000
MonthlyCharges      0.192858
PaperlessBilling    0.191454
SeniorCitizen       0.150541
PaymentMethod       0.107852
MultipleLines       0.038043
PhoneService        0.011691
gender             -0.008545
StreamingTV        -0.036303
StreamingMovies    -0.038802
InternetService    -0.047097
Partner            -0.149982
Dependents         -0.163128
DeviceProtection   -0.177883
OnlineBackup       -0.195290
TotalCharges       -0.199484
TechSupport        -0.282232
OnlineSecurity     -0.289050
tenure             -0.354049
Contract           -0.396150
Name: Churn, dtype: float64
<Figure size 1008x504 with 0 Axes>
In [41]:
X = df.drop(columns = ['Churn'])
y = df['Churn'].values
In [42]:
X_train, X_test, y_train, y_test = train_test_split(X,y,test_size = 0.30, random_state = 40, stratify=y)

Since the numerical features are distributed over different value ranges, I will use standard scalar to scale them down to the same range.

STANDARDIZATION

In [45]:
def distplot(feature, frame, color='r'):
    plt.figure(figsize=(8,3))
    plt.title("Distribution for {}".format(feature))
    ax = sns.distplot(frame[feature], color= color)
In [46]:
num_cols = ["tenure", 'MonthlyCharges', 'TotalCharges']
for feat in num_cols: distplot(feat, df)
In [47]:
df_std = pd.DataFrame(StandardScaler().fit_transform(df[num_cols].astype('float64')),
                       columns=num_cols)
for feat in numerical_cols: distplot(feat, df_std, color='c')
In [48]:
# Divide the columns into 3 categories, one ofor standardisation, one for label encoding and one for one hot encoding

# those that need one-hot encoding
cat_cols_ohe =['PaymentMethod', 'Contract', 'InternetService'] 

#those that need label encoding
cat_cols_le = list(set(X_train.columns)- set(num_cols) - set(cat_cols_ohe))
In [49]:
scaler= StandardScaler()

X_train[num_cols] = scaler.fit_transform(X_train[num_cols])
X_test[num_cols] = scaler.transform(X_test[num_cols])

MODELING

RANDOM FOREST CLASSIFIER

In [51]:
from sklearn import metrics
In [52]:
model_rf = RandomForestClassifier(n_estimators=500 , oob_score = True, n_jobs = -1,
                                  random_state =50, max_features = "auto",
                                  max_leaf_nodes = 30)
model_rf.fit(X_train, y_train)

# Make predictions
prediction_test = model_rf.predict(X_test)
print (metrics.accuracy_score(y_test, prediction_test))

0.8137440758293839
In [54]:
plt.figure(figsize=(4,3))
sns.heatmap(confusion_matrix(y_test, prediction_test),
                annot=True,fmt = "d",linecolor="k",linewidths=3)
    
plt.title(" RANDOM FOREST CONFUSION MATRIX",fontsize=14)
plt.show()
In [55]:
y_rfpred_prob = model_rf.predict_proba(X_test)[:,1]
fpr_rf, tpr_rf, thresholds = roc_curve(y_test, y_rfpred_prob)
plt.plot([0, 1], [0, 1], 'k--' )
plt.plot(fpr_rf, tpr_rf, label='Random Forest',color = "r")
plt.xlabel('False Positive Rate')
plt.ylabel('True Positive Rate')
plt.title('Random Forest ROC Curve',fontsize=16)
plt.show();

LOGISTIC REGRESSION

In [56]:
lr_model = LogisticRegression()
lr_model.fit(X_train,y_train)
accuracy_lr = lr_model.score(X_test,y_test)
print("Logistic Regression accuracy is :",accuracy_lr)

Logistic Regression accuracy is : 0.8090047393364929
In [58]:
lr_pred= lr_model.predict(X_test)
plt.figure(figsize=(4,3))
sns.heatmap(confusion_matrix(y_test, lr_pred),
                annot=True,fmt = "d",linecolor="k",linewidths=3)
    
plt.title("LOGISTIC REGRESSION CONFUSION MATRIX",fontsize=14)
plt.show()
In [59]:
y_pred_prob = lr_model.predict_proba(X_test)[:,1]
fpr, tpr, thresholds = roc_curve(y_test, y_pred_prob)
plt.plot([0, 1], [0, 1], 'k--' )
plt.plot(fpr, tpr, label='Logistic Regression',color = "r")
plt.xlabel('False Positive Rate')
plt.ylabel('True Positive Rate')
plt.title('Logistic Regression ROC Curve',fontsize=16)
plt.show();

DECISION TREE CLASSIFIER

In [60]:
dt_model = DecisionTreeClassifier()
dt_model.fit(X_train,y_train)
predictdt_y = dt_model.predict(X_test)
accuracy_dt = dt_model.score(X_test,y_test)
print("Decision Tree accuracy is :",accuracy_dt)

Decision Tree accuracy is : 0.7270142180094786

VOTING THE CLASSIFIER

Let's now forecast and rate the final model which is based on the largest majority of votes.

In [62]:
from sklearn.ensemble import VotingClassifier
clf1 = DecisionTreeClassifier()
clf2 = LogisticRegression()
clf3 = RandomForestClassifier()
eclf1 = VotingClassifier(estimators=[('gbc', clf1), ('lr', clf2), ('abc', clf3)], voting='soft')
eclf1.fit(X_train, y_train)
predictions = eclf1.predict(X_test)
print("Final Accuracy Score ")
print(accuracy_score(y_test, predictions))

Final Accuracy Score 
0.7630331753554502
In [63]:
plt.figure(figsize=(4,3))
sns.heatmap(confusion_matrix(y_test, predictions),
                annot=True,fmt = "d",linecolor="k",linewidths=3)
    
plt.title("FINAL CONFUSION MATRIX",fontsize=14)
plt.show()

According to the contingency table, there are a maximum of 1400+149=1549 real non-churn values, with the algorithms predicting 1400 as non-churn and 149 as churn. While there are 237+324=561 exact churn numbers, the algorithm to predict 237 of them to be non-churn and 324 to be churn.

Customer turnover has a negative impact on a company's profitability. To reduce customer turnover, a variety of techniques can be used. Customer turnover may be avoided by a firm getting to know its consumers. This entails identifying consumers who are likely to leave and attempting to enhance their satisfaction. Improving customer service is, of course, a key concern in addressing this problem. Another way to decrease client turnover is to build customer loyalty through meaningful experiences and customized service. Some businesses conduct surveys of existing customers to learn why they left in order to take proactive measures to prevent future customer churn.

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